# Why is $\lim_{x \to \infty} \arctan x$ defined when $\tan x$ is not defined at $x = \frac \pi 2?$

If one is undefined, shouldn't the other be undefined, too? They are inverse functions. For instance, since we have that

$$\lim_{x \to \frac{\pi}{2}} \tan x = \text{undefined}$$

so too should $$\lim_{x \to \infty} \arctan x = \text{undefined}$$

since there is no value for which $$\tan x$$ takes infinity.

• Where do you see it defined? May 19 '20 at 16:08
• how do you mean? May 19 '20 at 16:08
• You say "limit of $\arctan(x)$at x$=\infty$ defined" where do you see that? May 19 '20 at 16:11
• The confusion may be due to how $\infty$ is used in limits. It might look as if we are treating $\infty$ as a number but limits involving $\infty$ have their own definitions which don't actually involve $\infty$. It is just a suggestive shorthand. May 19 '20 at 16:13
• The points at which $\tan$ is undefined correspond to vertical asymptotes. When you flip the graph to get the inverse, these vertical asymptotes become horizontal.
– amd
May 19 '20 at 22:26

The limit of tangent at $$\frac\pi2$$ is undefined, but the limits approaching from below and above are defined: $$\lim_{x\rightarrow\frac\pi2,\;x<\frac\pi2}\tan(x)=+\infty, \\ \lim_{x\rightarrow\frac\pi2,\;x>\frac\pi2}\tan(x)=-\infty.$$ In particular, if we restrict the function $$\tan$$ to the open interval $$(-\frac\pi2,\frac\pi2)$$, then $$\tan$$ has limits at both ends.
The important thing to know is that $$\arctan$$ is defined as the inverse of this restricted function, not the general function on $$\Bbb R$$ (which is not even bijective). Therefore, nothing prevents $$\arctan$$ from having limits at $$-\infty$$ and $$+\infty$$, and in fact $$\lim_{x\rightarrow+\infty}\arctan(x)=\frac\pi2, \\ \lim_{x\rightarrow-\infty}\arctan(x)=-\frac\pi2.$$